KNEC EMFT and Communication Systems — June/July 2024 — Questions and Answers

Answer: The force between two stationary point charges is directly proportional to the product of their charges and inversely proportional to the square of the distance between them. It acts along the line joining the charges:

F = (1/(4πɛ)) (q₁q₂/R²) aR.

Like charges repel and unlike charges attract.

Answer:

Let P = (0, 0, 5).

From charge 1: R₁ = 2ax − ay + 5az, and from charge 2: R₂ = −2ax − ay + 5az. Both distances are R = √30 m.

Using k = 1/(4πε₀) = 8.99 × 10⁹:

E = kq(R₁ + R₂)/R³

E = 8.99 × 10⁹ × 25 × 10⁻⁶ (−2ay + 10az)/(√30)³

(i) E ≈ (−2.74 × 10³ay + 1.37 × 10⁴az) N/C.

For q₀ = 20 μC, F = q₀E.

(ii) F ≈ (−5.47 × 10⁻²ay + 2.74 × 10⁻¹az) N.

Answer: Rectangular/cartesian, cylindrical and spherical co-ordinate systems.

Answer:

μ = μ₀μᵣ = 4π × 10⁻⁷ × 5 H/m.

(i) B = μH = μ₀μᵣH = 4π × 10⁻⁷ × 5 × 12 ≈ 7.54 × 10⁻⁵ T.

(ii) For a linear magnetic material, M = (μᵣ − 1)H = (5 − 1)12 = 48 A/m.

Answer: The incremental magnetic field strength due to a current element is

dH = (I dℓ × aR)/(4πR²).

The incremental magnetic flux density is dB = μ dH.

Answer:

R₁₂ = P₂ − P₁ = −4ax + 3ay, so |R₁₂| = 5 m.

(I) aR1,2 = (−4ax + 3ay)/5 = −0.8ax + 0.6ay.

With I dℓ = 2π × 10⁻⁶ az A·m:

dH = [2π × 10⁻⁶ az × (−0.8ax + 0.6ay)]/(4π × 5²)

(II) dH = (−1.20 × 10⁻⁸ax − 1.60 × 10⁻⁸ay) A/m.

Answer:

1. ∮ E · dℓ = − d/dt ∫S B · dS (Faraday’s law).
2. ∮ H · dℓ = ∫S J · dS + d/dt ∫S D · dS (Ampere–Maxwell law).
3. ∮S D · dS = ∫V ρv dv (Gauss’s electric law).
4. ∮S B · dS = 0 (Gauss’s magnetic law).

Answer: Any five suitable properties are:

1. High electrical resistivity or very low conductivity.
2. High dielectric strength.
3. A suitable relative permittivity for the intended application.
4. Low dielectric loss or low loss tangent.
5. Good thermal stability and low moisture absorption.

Answer:

Comparing the phase with ωt + βz, β = 8 rad/m. The plus sign means that the wave travels in the negative z-direction.

(i) In free space, ω = βc = 8 × 3 × 10⁸ = 2.40 × 10⁹ rad/s.

(ii) λ = 2π/β = 2π/8 = 0.785 m.

(iii) With η₀ = 120π Ω and propagation direction −az:

H = (1/η₀)(−az × 48ax)cos(ωt + 8z)

H = −0.1273 cos(ωt + 8z)ay A/m.

(iv) Sav = E₀²/(2η₀) = 48²/(2 × 120π) ≈ 3.06 W/m², directed in the −az direction.

Answer:

• Tangential electric field: an × (E₂ − E₁) = 0.
• Tangential magnetic field: an × (H₂ − H₁) = Js.
• Normal magnetic flux density: an · (B₂ − B₁) = 0.
• Normal electric flux density: an · (D₂ − D₁) = ρs.

Answer:

For the calculation, take the interface as the x–y plane, with the normal in the z-direction, and assume no free surface current. The supplied paper gives the magnetic-flux-density vector and the figure’s geometry; the angle convention is therefore stated explicitly below.

|B₁| = √(3² + 1.8² + 1.2²) = 3.6986 T.

The angle made by the supplied vector with the normal is:

θ₁ = cos⁻¹(B₁z/|B₁|) = cos⁻¹(1.2/3.6986) ≈ 71.07°.

If the θ₁ label is read from the interface, as drawn in the scan, its complementary surface angle is 90° − 71.07° = 18.93°. Conventional incidence angles are normally measured from the normal.

At a source-free boundary, the normal component of B is continuous and the tangential component of H is continuous:

B₂n = B₁n = 1.2az T, and B₂t/μ₂ = B₁t/μ₁.

Therefore, using μr1 = 1 and μr2 = 8:

B₂ = 8(3ax + 1.8ay) + 1.2az

B₂ = (24a_x + 14.4a_y + 1.2a_z) T.

Note: B alone does not uniquely determine a wave’s Poynting-vector direction without the polarization/propagation convention. The two angle conventions above make the interpretation transparent for revision.

Answer:

Sources: accelerating electric charges; time-varying currents or atomic/electronic transitions.

Properties in free space: E and H are mutually perpendicular and both are perpendicular to the direction of propagation; the waves travel at approximately 3 × 10⁸ m/s. Other valid properties include E/H = η₀ and no attenuation in ideal free space.

Answer:

1. Digital signal processing gives improved detection, range/velocity measurement and clutter rejection.
2. Digital systems have better repeatability, noise immunity and storage of radar data.
3. They are programmable and can support automatic tracking, display and multifunction operation.

Answer:

A copyable labelled block diagram is:

[RF oscillator] ──┬──> [RF power amplifier] ──> [Tx antenna] ──> Moving target
                   │                                      ↑ echo
                   └──> [Reference/local oscillator]      [Rx antenna]
                                                          │
                                                          v
                                             [RF amplifier] ─> [Mixer]
                                                                  │
                                                                  v
                                                        [IF/filter amplifier]
                                                                  │
                                                                  v
                                                        [Doppler detector]
                                                                  │
                                                                  v
                                                            [Display]

The oscillator produces a continuous carrier. A portion is amplified and radiated toward the target while another portion is retained as the mixer reference. A moving target returns an echo shifted by the Doppler frequency. The received echo is mixed with the reference, producing a beat frequency fd = 2v/λ for a radial target velocity v. The filtered Doppler signal is detected, processed and displayed as target speed/direction information.

Answer:

λ = c/f = (3 × 10⁸)/(900 × 10⁶) = 0.3333 m.

For a monostatic radar, G = 4πAe/λ² and Pr = PtG²λ²σ/[(4π)³R⁴]. Hence:

Pt = Pr 4πR⁴λ²/(Ae²σ)

(i) With Pr = 350 × 10⁻¹² W, R = 25 × 10³ m, Ae = 7 m² and σ = 18 m²:

Pt ≈ 2.16 × 10⁵ W = 216.4 kW.

(ii) Received power varies as 1/R⁴:

Pr40 = 350(25/40)⁴ pW ≈ 53.4 pW.

Answer:

1. Amplitude gain: the ratio of output signal amplitude to input signal amplitude, Av = Vout/Vin.
2. Bandwidth: the range of frequencies occupied by the transmitted signal. For conventional AM with maximum modulating frequency fm, BW = 2fm.

Answer:

[Microphone/audio source] ─> [AF voltage amplifier] ─> [AF power amplifier]
                                                               │
                                                               v
[RF crystal oscillator] ─> [RF buffer/driver] ─────────> [High-level RF power amplifier/
                                                          collector modulator]
                                                               │
                                                               v
                                                        [Output filter]
                                                               │
                                                               v
                                                            [Antenna]

In high-level modulation, the audio power amplifier supplies the modulation power to the final RF power amplifier, where the carrier amplitude is varied by the audio signal.

Answer:

At 90 MHz, C90 = 200 + 50 = 250 pF. For resonance, f = 1/(2π√(LC)).

(i) L = 1/[(2π × 90 × 10⁶)²(250 × 10⁻¹²)] ≈ 1.25 × 10⁻⁸ H = 12.51 nH.

At 96 MHz:

C96,total = 1/[(2π × 96 × 10⁶)²L] ≈ 219.73 pF.

The new varactor capacitance is Cv,96 = 219.73 − 200 ≈ 19.73 pF.

(ii) Change in diode capacitance = 50 − 19.73 ≈ 30.27 pF decrease.

Answer:

For AM, channel bandwidth is 2B = 2 × 5 = 10 kHz, so the number of channels is 100/10 = 10 channels.

For SSB, channel bandwidth is B = 5 kHz, so the number of channels is 100/5 = 20 channels.

Answer:

1. VSB occupies less bandwidth and requires less transmitted power than DSB-AM while retaining the information needed for low-frequency components.
2. It is easier to generate and filter than a pure SSB signal, while reducing adjacent-channel interference compared with DSB-AM.

Answer:

1. Azimuth angle: the horizontal bearing of a satellite measured from a reference direction, normally true north, at the earth station.
2. Apogee: the point in an elliptical orbit farthest from the earth.
3. Perigee: the point in an elliptical orbit nearest to the earth.

Answer:

TRANSMIT PATH
[Baseband input] -> [MUX/codec] -> [Modulator] -> [IF amplifier]
                  -> [Up-converter] -> [High-power amplifier]
                  -> [Feed/earth-station antenna] -> Satellite

RECEIVE PATH
Satellite -> [Earth-station antenna/feed] -> [LNA/LNB]
          -> [Down-converter] -> [IF amplifier] -> [Demodulator]
          -> [De-mux/decoder] -> [Baseband output]

Answer:

Ku-band: approximately 12–18 GHz, commonly using about 12 GHz downlink and 14 GHz uplink.

C-band: approximately 4–8 GHz, commonly using about 4 GHz downlink and 6 GHz uplink.

Answer:

λ = c/f = (3 × 10⁸)/(9 × 10⁹) = 0.0333 m.

Free-space path loss:

Lfs(dB) = 20 log₁₀(4πR/λ), with R = 36,000 km = 3.6 × 10⁷ m.

(i) λ ≈ 3.33 × 10⁻² m.

(iii) Lfs ≈ 202.65 dB.

Using Friis’ equation in dB, Pr = Pt + Gt + Gr − Lfs:

Gr = 10log₁₀(2.8 × 10⁻⁶/4000) + 202.65 − 48

(ii) Gr ≈ 63.10 dB.

Answer:

1. Monochromaticity: the purity of a colour, or how nearly it consists of one dominant wavelength.
2. Brightness: the perceived intensity or light level of the displayed picture.
3. Hue: the basic colour or dominant wavelength, such as red, green or blue.

Answer:

1. Sound but no video: a fault in the video IF/detector, video amplifier, picture tube/display or video supply.
2. Flicker: incorrect or unstable vertical scanning frequency, weak power supply or a fault in the vertical oscillator/synchronising circuit.
3. Vertical lines: a horizontal deflection circuit, horizontal output stage, deflection coil or synchronisation fault.
4. Power ON but no picture or sound: a common power-supply, tuner/IF or system-control fault, or no usable aerial/RF input.

Answer:

[Camera 1] ─┐
[Camera 2] ─┼──> [Digital video recorder] ──> [Router] )))  [Wireless laptop]
[Camera 3] ─┘             │
                           └──> [Local monitor/storage, if fitted]

The three cameras send video to the DVR for recording and network streaming. The router provides the wireless link used by the laptop for viewing or administration.

Answer:

(i) fH = 625 × 30 = 18,750 Hz = 18.75 kHz.

(ii) TH = 1/fH = 1/18,750 ≈ 53.33 μs.

(iii) Aspect ratio = width/height = 24/18 = 4:3.

Answer:

1. HLR: the permanent central database containing the subscriber’s identity, service profile, current location information and routing details.
2. VLR: a temporary database serving a location area. It stores information about subscribers currently visiting that area and supports call setup, often using a temporary mobile subscriber identity.
3. AuC: the authentication centre. It stores security information and generates authentication and ciphering parameters used to verify the SIM/subscriber and protect communication.

Answer:

 [Electron gun/cathode] ===== electron beam =====> [Collector]
          │                         │
          v                         v
   [Focusing magnets]       [Helix slow-wave structure]
                                    ↑ RF input
                                    │
                              RF output/coupler

The electron gun produces a focused electron beam. The RF input travels along the helix, which slows the axial phase velocity so that it is close to the electron-beam velocity. The alternating RF field causes electron bunching and the bunched electrons transfer kinetic energy to the travelling RF wave. The amplified RF signal is taken from the output coupler, while the spent electrons are collected at the collector. Focusing magnets keep the beam aligned.

Answer:

Radius a = 0.065/2 = 0.0325 m, and operating wavelength λ = c/f = 0.03333 m.

(i) λc = 2πa/Kc = 2π(0.0325)/1.84 ≈ 0.11098 m.

Thus fc = c/λc ≈ 2.70 GHz.

(ii) Guide/group wavelength:

λg = λ/√[1 − (fc/f)²] ≈ 0.03495 m.

(iii) For the TE mode:

ZTE = η₀/√[1 − (fc/f)²] = 120π/√[1 − (2.70/9)²] ≈ 395 Ω.